--- title: "AM vs FM; error probability of coherent ASK" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:am-fm-compare-ask-error" source: https://www.bctengineer.com/q/semester_6/communication_systems/am-fm-compare-ask-error site: BCT Engineering --- # AM vs FM; error probability of coherent ASK Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2080 Baisakh (CS II), 5+5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/am-fm-compare-ask-error - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Compare A.M and F.M in terms of power efficiency and system complexity. Calculate the error probability of coherent ASK. ## Short answer AM has variable transmitted power, requires linear amplifiers, FOM ≤ 1/3, and simple envelope detection. FM has constant power, allows efficient class C amplifiers, FOM = 1.5β², but needs wider bandwidth and complex limiter/discriminator receivers. Coherent ASK error probability is P_e = ½erfc(√(E_b/2N_0)) = Q(√(E_b/N_0)), identical to BFSK and 3 dB worse than BPSK. ## Answer #### AM vs FM: power efficiency and complexity | Point | AM (DSB-FC) | FM | |---|---|---| | Transmitted power | Varies with modulation, $P_c(1+\mu^2/2)$ | Constant, $A_c^2/2$, independent of modulation | | Useful (sideband) power | At most 1/3 of total for a single tone ($\mu=1$) | All power is useful; carrier power is redistributed to sidebands | | Power amplifier | Linear (class A/B) amplifiers needed after low-level modulation, lower efficiency (class C only in high-level modulation) | Class C (non-linear), high efficiency, since envelope is constant | | Noise performance | FOM $\le 1/3$ for a single tone | FOM $=\frac32\beta^2$; much better for $\beta>0.5$ | | Bandwidth | $2W$ | $2(\beta+1)W$ (Carson), much wider | | Transmitter complexity | Simple (high-level collector/plate modulation) | More complex (VCO, AFC/frequency stabilisation, multipliers) | | Receiver complexity | Very simple (diode envelope detector) | Needs limiter, discriminator/PLL, de-emphasis | | Noise immunity | Amplitude noise directly affects output | Limiter removes amplitude noise; capture effect | So FM is **more power efficient** (better SNR per watt and efficient class C amplifiers) but uses **more bandwidth and more complex circuits**; AM is simpler and cheaper but wastes power in the carrier. #### Error probability of coherent ASK **ASK (on–off keying) signals** over $0\le t\le T_b$: $$ s_1(t)=A\cos\omega_ct\ \ (\text{bit }1),\qquad s_2(t)=0\ \ (\text{bit }0) $$ **Coherent receiver (correlator = matched filter):** ```text r(t) -->(x)--> integrate --> sample --> y > lam ? 1 : 0 ^ 0..Tb t=Tb | cos(wc t) (local carrier, in phase) ``` Correlator output at $t=T_b$, with $r(t)=s_i(t)+w(t)$: $$ y=\int_0^{T_b}r(t)\cos\omega_ct\,dt=a_i+n $$ - Bit 1: $a_1=\int_0^{T_b}A\cos^2\omega_ct\,dt=\frac{AT_b}{2}$ (for $f_c$ an integer multiple of $1/T_b$). - Bit 0: $a_2=0$. - Noise: $n=\int_0^{T_b}w(t)\cos\omega_ct\,dt$ is Gaussian, mean 0, variance $$ \sigma^2=\int_0^{T_b}\!\!\int_0^{T_b}\frac{N_0}{2}\delta(t-u)\cos\omega_ct\cos\omega_cu\,dt\,du=\frac{N_0}{2}\cdot\frac{T_b}{2}=\frac{N_0T_b}{4} $$ Threshold midway: $\lambda=\frac{AT_b}{4}$. Using the binary result $P_e=\frac12\operatorname{erfc}\left(\frac{a_1-a_2}{2\sqrt2\sigma}\right)$: $$ \begin{aligned} \frac{a_1-a_2}{2\sqrt2\,\sigma} &= \frac{AT_b/2}{2\sqrt2\sqrt{N_0T_b/4}}=\frac{A\sqrt{T_b}}{2\sqrt{2N_0}}=\sqrt{\frac{A^2T_b}{8N_0}} \end{aligned} $$ Energy of a "1" pulse is $E_1=\frac{A^2T_b}{2}$; the average energy per bit (half the bits are 0) is $E_b=\frac{E_1}{2}=\frac{A^2T_b}{4}$. Hence $\frac{A^2T_b}{8N_0}=\frac{E_b}{2N_0}$ and $$ P_e(\text{ASK})=\frac12\operatorname{erfc}\left(\sqrt{\frac{E_b}{2N_0}}\right)=Q\left(\sqrt{\frac{E_b}{N_0}}\right) $$ (In terms of peak energy $E_1$: $P_e=\frac12\operatorname{erfc}\sqrt{E_1/4N_0}$.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can FM use a more efficient power amplifier than AM? A. FM has a smaller bandwidth B. FM has constant envelope, so a non-linear class C amplifier can be used C. FM carries no carrier D. FM needs a limiter at the transmitter Answer: B. Constant transmitted power with constant envelope allows class C; AM needs a linear amplifier. (Key point: Power amplifier type) 2. What fraction of total power in AM (DSB-FC) at most carries information (mu = 1)? A. 1/2 B. 2/3 C. 1/3 D. 1/4 Answer: C. Sideband power is mu^2/(2+mu^2) of the total = 1/3 at mu = 1. (Key point: AM sideband power) 3. In the coherent ASK correlator, what is the noise variance sigma^2? A. N0 Tb / 4 B. N0 Tb / 2 C. N0 Tb D. N0 / 2 Answer: A. sigma^2 = (N0/2)(Tb/2) = N0 Tb/4 since the integral of cos^2 over Tb is Tb/2. (Key point: ASK noise variance) 4. What is Pe of coherent ASK in terms of average Eb? A. (1/2) erfc( sqrt(Eb/N0) ) B. (1/2) erfc( sqrt(Eb/2N0) ) = Q( sqrt(Eb/N0) ) C. (1/2) erfc( sqrt(2Eb/N0) ) D. (1/2) exp(-Eb/2N0) Answer: B. a1 - a2 = A Tb/2 gives argument sqrt(A^2Tb/8N0) = sqrt(Eb/2N0), with average Eb = A^2Tb/4. (Key point: ASK error probability) ## Close variants set in other exams - [AM vs FM efficiency and complexity; coherent ASK error](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-fm-bw-efficiency-ask-error) — Communication Systems, chapter 9, asked 2× ## Related in Communication Systems - [Error probability of coherent ASK](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aask-coherent-error-probability) — Communication Systems, chapter 9, asked 2× - [Coherent ASK and PSK error; ASK needs double power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aask-psk-error-double-power) — Communication Systems, chapter 9 - [Threshold effect in FM; error probability of coherent PSK](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Athreshold-effect-psk-error) — Communication Systems, chapter 9